It follows from the Theorem REF (i) that \(\operatorname{H}^4_{\mathfrak {m}}(M\otimes _RM) \ne 0\) . Therefore, we proceed to show the other claims of the example.
Consider the following exact sequences:
\((a):\ 0\rightarrow \Omega ^2k\rightarrow R^{\beta _1}\rightarrow \Omega ^1k\rightarrow 0, \ \ \ (b):\ 0\rightarrow \Omega ^1k\rightarrow R\rightarrow k\rightarrow 0.\) 
The exact sequence (a) induces the following exact sequence:
\(0\rightarrow \operatorname{Tor}_1^R(\Omega ^1k,\Omega ^2k)\rightarrow \Omega ^2k\otimes _R\Omega ^2k\rightarrow R^{\beta _1}\otimes _R\Omega ^2k\rightarrow \Omega ^1k\otimes _R\Omega ^2k\rightarrow 0.\) 
We break it down into
\( (a.1):\ 0\rightarrow \operatorname{Tor}_1^R(\Omega ^1k,\Omega ^2k)\rightarrow \Omega ^2k\otimes _R\Omega ^2k\rightarrow X\rightarrow 0,\) 
\( (a.2):\ 0\rightarrow X\rightarrow R^{\beta _1}\otimes _R\Omega ^2k\rightarrow \Omega ^1k\otimes _R\Omega ^2k\rightarrow 0.\) 
It follows from \((a.2)\)  that \(\operatorname{depth}(X)>0\) . Hence the exact sequence \((a.1)\) 
induces the isomorphisms
\(\operatorname{H}^0_{\mathfrak {m}}(\operatorname{\Omega }^2k\otimes _R\operatorname{\Omega }^2k)\simeq \operatorname{H}^0_{\mathfrak {m}}(\operatorname{Tor}_1^R(\operatorname{\Omega }^1k,\operatorname{\Omega }^2k))\simeq \operatorname{Tor}_1^R(\operatorname{\Omega }^1k,\operatorname{\Omega }^2k)\simeq \operatorname{Tor}_4^R(k,k).\qquad \mathrm {(\ref {e}.1)}\) 
Therefore, \(\ell (\operatorname{H}^0_{\mathfrak {m}}(M\otimes _RM))=\beta _4(k)\) .
Since \(\operatorname{Tor}_1^R(\operatorname{\Omega }^1k,\operatorname{\Omega }^2k)\)  has finite length, \(\operatorname{H}^i_{\mathfrak {m}}(\operatorname{Tor}_1^R(\operatorname{\Omega }^1k,\operatorname{\Omega }^2k))=0\)  for all \(i>0\) . Thus, the exact sequence \((a.1)\)  induces the isomorphism
\(\operatorname{H}^j_{\mathfrak {m}}(\operatorname{\Omega }^2k\otimes _R\operatorname{\Omega }^2k)\simeq \operatorname{H}^j_{\mathfrak {m}}(X)\)  for all \(j>0\) .
Note that \(\operatorname{H}^0_{\mathfrak {m}}(\operatorname{\Omega }^2k)=\operatorname{H}^1_{\mathfrak {m}}(\operatorname{\Omega }^2k)=0\) . Therefore, the exact sequence \((a.2)\)  induces the exact sequence:
\(0=\operatorname{H}^0_{\mathfrak {m}}(R^{\beta _1}\otimes _R\operatorname{\Omega }^2k)\rightarrow \operatorname{H}^0_{\mathfrak {m}}(\operatorname{\Omega }^2k\otimes _R \operatorname{\Omega }^1k)\rightarrow \operatorname{H}^1_{\mathfrak {m}}(X)\rightarrow \operatorname{H}^1_{\mathfrak {m}}(R^{\beta _1}\otimes _R\operatorname{\Omega }^2k)=0.\) 
Hence, we get the following isomorphisms
\(\operatorname{H}^1_{\mathfrak {m}}(\operatorname{\Omega }^2k\otimes _R\operatorname{\Omega }^2k)\simeq \operatorname{H}^1_{\mathfrak {m}}(X)\simeq \operatorname{H}^0_{\mathfrak {m}}(\operatorname{\Omega }^2k\otimes _R \operatorname{\Omega }^1k).\qquad \mathrm {(\ref {e}.2)}\) 
The exact sequence \((b)\)  induces the following exact sequence:
\(0\longrightarrow \operatorname{Tor}_1^R(k,\operatorname{\Omega }^2k)\rightarrow \operatorname{\Omega }^1k\otimes _R\operatorname{\Omega }^2k \longrightarrow \operatorname{\Omega }^2k\rightarrow k\otimes _R\operatorname{\Omega }^2k\longrightarrow 0.\) 
We break it down into the following short exact sequences:
\((b.1):\ 0\rightarrow \operatorname{Tor}_1^R(k,\operatorname{\Omega }^2k)\rightarrow \operatorname{\Omega }^1k\otimes _R\operatorname{\Omega }^2k\rightarrow Y\rightarrow 0,\) 
\((b.2):\ 0\rightarrow Y\rightarrow \operatorname{\Omega }^2k\rightarrow k\otimes _R\operatorname{\Omega }^2k\rightarrow 0.\) 
It follows from \((b.2)\)  that \(\operatorname{depth}(Y)>0\) . Hence, the exact sequence \((b.1)\)  induces the isomorphism
\(\operatorname{H}^0_{\mathfrak {m}}(\operatorname{\Omega }^2k\otimes \operatorname{\Omega }^1k)\simeq \operatorname{H}^0_{\mathfrak {m}}(\operatorname{Tor}_1^R(k,\operatorname{\Omega }^2k))\simeq \operatorname{Tor}_1^R(k,\operatorname{\Omega }^2k)\simeq \operatorname{Tor}_3^R(k,k).\qquad \mathrm {(\ref {e}.3)}\) 
By (REF .2) and (REF .3), we have
\(\operatorname{H}^1_{\mathfrak {m}}(\operatorname{\Omega }^2k\otimes _R\operatorname{\Omega }^2k)\simeq \operatorname{H}^0_{\mathfrak {m}}(\operatorname{\Omega }^2k\otimes _R \operatorname{\Omega }^1k)\simeq \operatorname{Tor}_3^R(k,k).\qquad \mathrm {(\ref {e}.4)}\) 
Therefore, \(\ell (\operatorname{H}^1_{\mathfrak {m}}(M\otimes _RM))=\beta _3(k)\) .
Recall that \(\operatorname{H}^1_{\mathfrak {m}}(\operatorname{\Omega }^2k) =\operatorname{H}^3_{\mathfrak {m}}(\operatorname{\Omega }^2k)=0\)  and that \(\operatorname{H}^2_{\mathfrak {m}}(\operatorname{\Omega }^2k)=k\)  (see [1]}).
Therefore, applying the functor \(\Gamma _{\mathfrak {m}}(-)\)  to the exact sequence \((a.2)\) , we get the long exact sequence:
\(0\rightarrow \operatorname{H}^1_{\mathfrak {m}}(\operatorname{\Omega }^2k\otimes _R\operatorname{\Omega }^1k)\rightarrow \operatorname{H}^2_{\mathfrak {m}}(X)\rightarrow \operatorname{H}^2_{\mathfrak {m}}(R^{\beta _1}\otimes _R\operatorname{\Omega }^2k)\rightarrow \operatorname{H}^2_{\mathfrak {m}}(\operatorname{\Omega }^2k\otimes _R\operatorname{\Omega }^1k)\rightarrow \operatorname{H}^3_{\mathfrak {m}}(X)\rightarrow 0.\qquad \mathrm {(\ref {e}.5)}\) 
Also, the exact sequences \((b.1)\)  and \((b.2)\)  induce the following isomorphisms:
\(\operatorname{H}^i_{\mathfrak {m}}(\operatorname{\Omega }^2k\otimes _R\operatorname{\Omega }^1k)\simeq \operatorname{H}^i_{\mathfrak {m}}(Y)\ \text{and} \ \ \operatorname{H}^{j}_{\mathfrak {m}}(Y)\simeq \operatorname{H}^{j}_{\mathfrak {m}}(\operatorname{\Omega }^2k)\ \text{ for all } i>0\text{ and }\ j>1,\qquad \mathrm {(\ref {e}.6)}\) 
\(k^{\oplus \beta _3}\simeq k\otimes _R\operatorname{\Omega }^2k\simeq \operatorname{H}^0_{\mathfrak {m}}(k\otimes _R\operatorname{\Omega }^2k)\simeq \operatorname{H}^1_{\mathfrak {m}}(Y)\simeq \operatorname{H}^1_{\mathfrak {m}}(\operatorname{\Omega }^2k\otimes _R\operatorname{\Omega }^1k).\qquad \mathrm {(\ref {e}.7)}\) 
Therefore, by using (REF .6) and (REF .7) one can rewrite the exact sequence (REF .5) as follows:
\(0\rightarrow k^{\oplus \beta _3}\rightarrow \operatorname{H}^2_{\mathfrak {m}}(X)\rightarrow k^{\oplus \beta _1}\stackrel{f}{\rightarrow }k\stackrel{g}{\rightarrow }\operatorname{H}^3_{\mathfrak {m}}(X)\rightarrow 0.\qquad \mathrm {(\ref {e}.8)}\) 
Claim. \(\operatorname{H}^3_{\mathfrak {m}}(M\otimes M)\cong \operatorname{H}^3_{\mathfrak {m}}(X)\ne 0\) .
To establish the claim, note, by [2]}, we have the following isomorphisms:
\(\operatorname{H}^3_{\mathfrak {m}}(X)^v\simeq \operatorname{H}^3_{\mathfrak {m}}(M\otimes M)^v\simeq \operatorname{H}^2_{\mathfrak {m}}(M^\ast \otimes M^\ast )\simeq \operatorname{H}^2_{\mathfrak {m}}(\operatorname{\Omega }^3k\otimes \operatorname{\Omega }^3k),\qquad \mathrm {(\ref {e}.9)}\) 
where \((-)^v\)  is the Matlis dual.
Recall that \(\operatorname{depth}(\operatorname{\Omega }^3k)+\operatorname{depth}(\operatorname{\Omega }^3k)=64+2+1\) . In view of [3]} we see that \(\operatorname{H}^2_{\mathfrak {m}}(\operatorname{\Omega }^3k\otimes \operatorname{\Omega }^3k)\ne 0\) . Now the assertion follows from (REF .9) and the proof of the claim is complete.
Since \(\operatorname{H}^3_{\mathfrak {m}}(X)\)  is non-zero and a homomorphic image of \(k\) ,
we conclude that \(\operatorname{H}^3_{\mathfrak {m}}(X)\simeq k\) . In particular, \(\ell (\operatorname{H}^3_{\mathfrak {m}}(M\otimes M))=\ell (\operatorname{H}^3_{\mathfrak {m}}(X))=\beta _{0}(k)\) .
It follows from the exact sequence (REF .8) that \(g\)  is an isomorphism and \(f=0\) 
so that \(\ell (\operatorname{H}^2_{\mathfrak {m}}(M\otimes M))=\ell (\operatorname{H}^2_{\mathfrak {m}}(X))=\beta _3(k)+\beta _1(k)= 2 \beta _1(k)\) , as required.
